How can a linear ordinary differential equation with constant coefficients be solved using linear algebra?
Solution
Consider first a scalar equation of order \(n\), \[ y^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_1y'+a_0y=0. \] Introduce the state vector \[ Y=\begin{pmatrix}y\\y'\\\vdots\\y^{(n-1)}\end{pmatrix}. \] Then the equation is equivalent to the first-order system \[ Y'=AY, \] where \(A\) is the companion matrix \[ A= \begin{pmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&&&\ddots&\vdots\\ 0&0&0&\cdots&1\\ -a_0&-a_1&-a_2&\cdots&-a_{n-1} \end{pmatrix}. \]
The solution with initial condition \(Y(0)=Y_0\) is \[ Y(t)=e^{tA}Y_0. \] Thus the problem reduces to computing the matrix exponential.
If \(A\) is diagonalizable, say \[ A=PDP^{-1}, \] then \[ e^{tA}=Pe^{tD}P^{-1}, \] and the diagonal entries of \(e^{tD}\) are \(e^{\lambda_i t}\).
More generally, put \(A\) in Jordan form. For a Jordan block \[ J=\lambda I+N, \qquad N^r=0, \] one has \[ e^{tJ}=e^{\lambda t}e^{tN} =e^{\lambda t}\sum_{k=0}^{r-1}\frac{t^kN^k}{k!}. \] This recovers the usual solutions \(t^ke^{\lambda t}\) attached to repeated roots of the characteristic polynomial.